SOLUTION: prove that there do not exist three consecutive integer values of n for which 73n+70 is a perfect square.

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Question 1193263: prove that there do not exist three consecutive integer values of n for which 73n+70 is a perfect square.
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20064)   (Show Source): You can put this solution on YOUR website!
We can do better than that.  We can prove that there do not even exist 
TWO consecutive integer values of n for which 73n+70 is a perfect square.

Assume for contradiction that there are two consecutive integers
'a', and 'a+1', and also integers 'x' and 'y' such that 















73 is a prime number and only has factors 1 and 73, so



Adding those equations gives 2y = 74, or y  = 37
Substituting 37-x=1, -x=-36, x=36

Now we substitute for x and y in









So 'a' cannot be an integer.  And since we assumed that
'a' was an integer, we have reached a contradiction.

Since there are no TWO consecutive integer values for n
for which 73n+70 is a perfect square, there certainly are
no THREE such.

Edwin
Answer by ikleyn(52887)   (Show Source): You can put this solution on YOUR website!
.
prove that there do not exist three consecutive integer values of n for which 73n+70 is a perfect square.
~~~~~~~~~~~~~~~~


            To make a harmony from a disharmony,  I would re-formulate the problem this way 

            Prove that there do not exist three consecutive integer values 
            of n such that the numbers  73n+70  all are perfect squares.


Let assume that there are (do exist) three consecutive integer values of n such that
three corresponding numbers 73n+70 are perfect squares.

Consider remainders of division these three integer numbers by 3.
Notice that 73n+70 gives the same remainder modulo 3 as the number (n+1).

So, the remainders of division the numbers 73n+70 by 3 are three numbers 0, 1 and 2 in some order.

In this chain of arguments, the fact which is important for us is that the remainder 2 with necessity
is one of the three remainders of (73n+70) modulo 3.

But no one perfect square may have remainder 2 when divided by 3: the possible remainders are only numbers 0 or 1.

This contradiction PROVES the statement.

Solved.



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